MATH 136
The Area Function
Exercises
x
1. Let A(x) = ∫ t + 3 dt for x ≥ –3.
1
(a) Explain what A(x) measures. Show the result pictorially.
(b) Evaluate Aʹ (x) .
€ ) and explain the result.
(c) Evaluate lim A(x
x →1
(d) Find a function formula for A(x) .
(e) Use the function formula to evaluate A(6) , A(−2) , lim A(x ) ,
x→ ∞
lim
x → −3+
A( x) and
explain the results.
x
2. Let A(x) =
∫
8
2
dt for x > 0.
t 1/ 3
(a) Explain what A(x) measures. Show the result pictorially.
(b) Evaluate Aʹ (x) .
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(c) Evaluate lim A(x ) and explain the result.
x→ 8
(d) Find a function formula for A(x) .
(e) Use the function formula to evaluate A(64) , A(1) , lim A(x ) ,
x→ ∞
lim A(x ) and explain
x → 0+
the results.
x
3. Let A(x) =
36
x
5 / 2 dt for > 0.
t
4
(a) Explain what A(x) measures. Show the result pictorially.
(b) Evaluate Aʹ (x) .
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(c) Evaluate lim A(x ) and explain the result.
∫
x→ 4
(d) Find a function formula for A(x) .
(e) Use the function formula to evaluate A(16) , A(1) , lim A(x ) ,
x→ ∞
the results.
x
4. Let A(x) = ∫ e 2t dt for all x .
0
(a) Explain what A(x) measures. Show the result pictorially.
(b) Evaluate Aʹ (x) .
€ ) and explain the result.
(c) Evaluate lim A(x
x→ 0
(d) Evaluate lim A(x ) by inspection and explain the result.
x→ ∞
lim A(x ) and explain
x → 0+
Solutions
1. (a) For x ≥ –3, A(x) measures the area under the graph of
over the interval 1 to x . (b) Aʹ (x) = x + 3 .
t +3
A(x ) = 0, since the line segment obtained when x equals 1
(c)
€ xlim
→1
1 x
has no area.€ The accumulated area shrinks to 0 as x approaches
the starting point of 1.
€
(d) For x ≥ –3,
€
x
x
2
2
2
16
3/ 2
3/ 2 2
= (x + 3)
sq. units.
− (4)3/2 = (x + 3)3/ 2 −
A(x) = ∫ t + 3 dt = (t + 3)
3
3
3
3
3
1
1
€
2 3/ 2 16 38
(e) A(6) = (9) −
=
sq. units, which is the area under t + 3 from 1 to 6.
3
3
3
14
A(−2) = − ; when x = –2, the signed area from 1 to –2 is accumulated in reverse
3
direction from right to left and equals –14/3 sq. units.
2
16
lim A(x ) = €lim
( x + 3)3 /2 −
= +∞.
3
x→ ∞
x→ ∞ 3
from 1 to x also increases to ∞.
As x increases to ∞, the accumulated area
2
16€
16
( x + 3)3/2 −
= −
sq. units. As x decreases to –3, the
3
3
x€→ −3+
x → −3+ 3
accumulated signed area from 1 to x decreases to –16/3.
______________________________________________________________________________
€
A( x) =
lim
lim
2. (a) For x > 0 , A(x) measures the area under the graph
€
of 2 / t1 / 3 over [8, x ].
(b) Aʹ (x) = 2 / x1/ 3 .
A(x ) = 0, since the line segment obtained when
(c)
€ xlim
→8
x equals€8 has no area. The accumulated
area shrinks to
€
0 as x approaches the starting point of 8.
€
x
€
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(d) For x > 0 , A(x) =
2
3 t 2/3
x
= 3x 2/ 3 − 3(82/ 3 ) = 3x
8
x
2/ 3
− 12 sq. units.
8
t
8
2/ 3
(e) A(64) = 3 (64 ) − 12 = 36 sq. units, which is the area under 2 / t1 / 3 from 8 to 64.
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A(1) = –9 sq. units; which gives the accumulated area in reverse direction from 8 to 1.
lim A(x ) = lim (3x
∫
1/ 3 dt =
2/3
€
− 12) = +∞: As x increases to ∞, the accumulated area from 8
x→ ∞
x→ ∞
to x also increases to ∞.
lim A(x ) = –12: As x decreases to 0, the signed (right to left)
x → 0+
€ decreases to –12.
area from 8 to x remains finite and
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€
€
3.
(a) For x > 0 , A(x) measures the area under the graph of
(b) Aʹ (x) = 36 / x 5 / 2 .
36 / t 5/ 2 over the interval [4, x ].
€
lim A(x ) = 0 since the line segment obtained when x equals 4
(c) €
x→ 4
€
has no area. The accumulated
area shrinks€to 0 as x approaches
the starting point of 4.
x
€
−24 x
−24 −24
24
36
(d) For x > 0 , A(x) = ∫ 5 / 2 dt = 3 / 2 = 3 / 2 − 3 / 2 = 3 − 3/ 2 .
t
t
x€
4
x
4
4
x
4
24
= 2.625 sq. units, which is the area under 36 / t 5/ 2 from 4 to 16.
3/2
16
A(1) = –21 sq. units, which is the “backwards” area under the curve from 4 to 1.
€(e) A(16) = 3 −
⎛
24 ⎞
€ accumulated area from 4 to
lim A(x ) = lim ⎜ 3 − 3/ 2 ⎟ = 3. As x increases to ∞, the
x→ ∞
x → ∞⎝
x ⎠
x remains finite and increases to 3.
⎛
24 ⎞€
lim ⎜ 3 − 3 /2 ⎟ = –∞. As x decreases to 0, the signed (right to left) area
⎝
x ⎠
x → 0+
x → 0+
from 4 to x becomes larger and larger in magnitude, thus decreasing to –∞.
lim A(x ) =
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€
______________________________________________________________________________
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4. (a) A(x) measures the area under the graph of e 2t
over 0 to x .
(b) Aʹ (x) = e2 x .
(c) lim A(x ) = 0, since the line segment obtained
when
€
x→ 0
€
x equals 0 has no area.
(d) Obviously lim A(x ) = +∞. As x increases to ∞, the accumulated area from 0 to x
x→ ∞
€
also increases to ∞.
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